为什么这个闭包函数没有编译器错误？

Question

我正在阅读 Marijn Haverbeke 的优秀著作“Eloquent JavaScript”。 https://eloquentjavascript.net/

我不明白这个关于没有定义number闭包的例子，但没有错误。

number是函数还是参数？

没有任何迹象表明第二次将number作为参数传入。

function multiplier(factor) {
  return number => number * factor;
}

let twice = multiplier(2);
console.log(twice(5));
// → 10

Answer 1

理解闭包的工作原理已经够难了，但是当您正在查看的示例将函数声明与箭头函数任意混合时 - 就像这个一样 - 如果您不了解箭头函数的工作原理，它会使理解变得更加困难。

这是一个稍微简单的示例，它不使用箭头函数来显示正在发生的事情。

 // `multipler` takes a factor as an argument function multiplier(factor) { // It returns a function that - when it's called - // accepts a number return function (number) { // And the return from that function // is the factor * number return number * factor; } } // So we call `multipler` with a factor and assign the // function it returns to our `twice` variable let twice = multiplier(2); // We can then call the function assigned // to `twice` with a number, and the resulting // return from that function will be factor * number console.log(twice(5));

就使用该箭头函数的示例而言：

 // We pass in a factor to the `multipler` function multiplier(factor) { // We return a function that accepts a number // and returns factor * number // (I've added parentheses around the number // parameter to clearly show it) return (number) => number * factor; } // So we call `multipler` with a factor and assign the // function it returns to our `twice` variable let twice = multiplier(2); // We can then call the function assigned // to `twice` with a number, and the resulting // return from that function will be factor * number console.log(twice(5));